Question

Solenoid A has twice the diameter, three times the length, and four times the number of turns of solenoid B. The two solenoids have currents of equal magnitudes flowing through them. Find the ratio of the magnitude of the magnetic field in the interior of solenoid $A$ to that of solenoid $B$

Short answer

Answer: The ratio of the magnetic field inside solenoid A to the magnetic field inside solenoid B is 4:3.

Step-by-step solution

Recall the formula for the magnetic field inside a solenoid

The magnetic field B inside a solenoid is given by the formula: B = μ₀ * n * I where μ₀ is the permeability of free space (μ₀ ≈ 4π × 10^(-7) Tm/A), n is the number of turns per unit length (turns/m), and I is the current (in A).

Find the number of turns per unit length of solenoids A and B

We are given that solenoid A has 4 times the number of turns as solenoid B. Let n_B be the number of turns per unit length of solenoid B, so the number of turns per unit length for solenoid A will be 4n_B. We also know that solenoid A is three times longer than solenoid B, so the length of solenoid A is 3 times the length of solenoid B. Let l_B be the length of solenoid B, then the length of solenoid A will be 3l_B. Now we can find the number of turns per unit length for solenoids A and B: n_A = (4n_B) / (3l_B) = (4/3)n_B n_B = n_B

Find the ratio of magnetic fields

Now we can use the formula for magnetic fields inside solenoids to find the ratio of magnetic fields for solenoids A and B: B_A / B_B = (μ₀ * n_A * I) / (μ₀ * n_B * I) Since the currents in both solenoids have equal magnitudes, the current factor in the above equation cancels out: B_A / B_B = n_A / n_B Now we can substitute the values we found for n_A and n_B in step 2: B_A / B_B = ((4/3)n_B) / n_B The n_B factors cancel out, leaving us with: B_A / B_B = 4/3 So, the ratio of the magnitude of the magnetic field in the interior of solenoid A to that of solenoid B is 4:3.

Key concepts

Solenoid Properties

Solenoids are devices widely used in electrical engineering. They generate a uniform magnetic field inside a coil when an electric current passes through it. The three key properties of a solenoid include its diameter, length, and the number of turns (coils). These properties affect the behavior and magnitude of the magnetic field within the solenoid.

• **Diameter**: Impacts the cross-sectional area but does not directly influence the magnetic field inside an ideal long solenoid.
• **Length**: A longer solenoid allows a more uniform and stronger magnetic field assuming constant current density.
• **Number of Turns**: More turns increase the magnetic field since it effectively brings more magnetic loops per unit length.

Different configurations of these properties can drastically change how the solenoid behaves, particularly when comparing solenoids A and B in this scenario. Solenoid A has a larger diameter, is longer, and has more turns compared to solenoid B. However, the magnetic field inside is independent of the solenoid’s diameter.

Magnetic Field Formula

The magnetic field inside a solenoid is determined by the equation:
$$B={\mu }_0\cdot n\cdot I$$
Here, $B$ represents the magnetic field strength, ${\mu }_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current passing through the solenoid. This formula is crucial for understanding how changes in various parameters affect the solenoid's magnetic field.

• **Permeability of Free Space (${\mu }_0$)**: A constant that relates magnetic field and magnetic force in vacuum, approximately equal to $4\pi \times {10}^{-7}\text{T m/A}$.
• **Number of Turns Per Unit Length ($n$)**: A major factor influencing the field's strength; as the number of turns increase, so does the field strength.
• **Current ($I$)**: The magnetic field strength is directly proportional to the current flowing through the solenoid.

By utilizing this formula, we can deduce the behavior of magnetic fields in solenoids of varying designs.

Permeability of Free Space

The permeability of free space, denoted as ${\mu }_0$, is a fundamental constant in electromagnetism. It specifies the measure of ability for vacuum to sustain a magnetic field. This constant is one of the building blocks of the magnetic field formula of a solenoid.

• **Value**: ${\mu }_0$ is approximately $4\pi \times {10}^{-7}\text{ T m/A}$.
• **Role**: Determines the magnitude of interaction between the magnetic field and the current in vacuum.
• **Environment**: In real-life scenarios, the value can change slightly in different materials; however, ${\mu }_0$ is used for calculations in a vacuum.

The permeability of free space allows us to simplify many electromagnetic calculations, especially within the context of solenoids where it helps determine the baseline magnetic field in the absence of any material medium.

Ratio of Magnetic Fields

When comparing two solenoids with different physical properties, we often want to determine the ratio of their magnetic fields. This calculation helps understand how modifications, such as changes in the number of coils or length, affect the magnetic field strength.
In the given exercise, we calculated the ratio of the magnetic fields of solenoids A and B using the formula:
$$\frac{B_A}{B_B}=\frac{n_A}{n_B}$$
Here, $n_A$ and $n_B$ represent the number of turns per length for solenoids A and B respectively. The current, permeability of free space, and other constants were considered equal, simplifying the ratio to depend solely on $n$.
By utilizing solenoid A's properties (four times the number of turns and three times the length of solenoid B), we derive that the magnetic field ratio $B_A/B_B$ is $4/3$. This insight shows how crucially the configuration of turns and length alter the magnetic behavior.